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Partial fractions integral problem

  1. Jul 12, 2006 #1
    Hi, IM trying to evaluate this, and I cant get started..I tried integration by partial fractions and substitution but I keep getting stuck.

    [tex] \int_0^2 \frac{x-3}{2x-3}dx[/tex]

    Any hints would help, Thanks
  2. jcsd
  3. Jul 12, 2006 #2


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    It should be straightforward using partial fractions. Remember, since the polynomial in the numerator has the same degree as the one in the denominator, you need to add an extra unknown constant:

    [tex]\frac{x-3}{2x-3} = \frac{A}{2x-3} +B [/tex]

    Or, by inspection, just write x-3=(x-3/2)-3/2=1/2(2x-3)-3/2.
  4. Jul 12, 2006 #3
    Consider that [tex]\frac{x-3}{2x-3}[/tex] has a vertical asymptote.
  5. Jul 12, 2006 #4


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    You can also do long (or synthetic) division to get

  6. Jul 12, 2006 #5
    oh ok thanks..I had part of that earlier..but the last step makes it work

  7. Jul 12, 2006 #6
    yes I got [tex]\frac{x-3}{2x-3}=\frac{1}{2}-\frac{3}{2(2x-3)}[/tex]

    ..and continue to get [tex]-3ln(4x-6) + \frac{x}{2} \right]_2^0[/tex]

    =[tex] [-3ln2 +1] - [-3ln|-6|][/tex]
  8. Jul 12, 2006 #7
    I think that what durt mentioned should be reemphasized that this is not just a definite integral but an improper integral since the function you are integrating has a vertical assymptote at x = 2/3 which lies inside of the interval you are integrating over.
  9. Jul 14, 2006 #8
    But it's not continuous as D_leet said so...I'd break it in two integrals...

    remember that integral from a to b + inegral from b to c = integral from a to c. Because of the asymptote, it has a sign change in there also...it goes from + to - so that complicates the one-integral process.
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